Answer:
A particle
Explanation:
Modern quantum theory holds that light has both wave-like and particle-like properties. When the length scales involved are large compared to the wavelengths of light (ex., forming images with thin lenses), the
particle nature of light dominates.
Im not sure but i think its 350. I really hope im right
Answer:
Check the explanation
Explanation:
AT = A0 e(-T/H)
... where A0 is the starting activity, AT is the activity at some time T, and H is the half-life, in units of T.
Substituting what we know, we get...
0.71 = (1) e(-T/5730)
Solve for T...
loge(0.71) = -T/5730
T = -loge(0.71)(5730)
T = 1962 (conservatively rounded, T = 2000)
similarly for all
for aboriginal charcoal
0.28 = (1) e(-T/5730)
Solve for T...
loge(0.28) = -T/5730
T = -loge(0.28)(5730)
T = 7294 (conservatively rounded, T = 7000)
for mayan headdress
0.89 = (1) e(-T/5730)
Solve for T...
loge(0.89) = -T/5730
T = -loge(0.89)(5730)
T = 667 (conservatively rounded, T = 700)
for neanderthal
0.05 = (1) e(-T/5730)
Solve for T...
loge(0.05) = -T/5730
T = -loge(0.05)(5730)
T = 17165 (conservatively rounded, T = 17000)
Cycloalkanes are those saturated organic compounds which exist in the form of Rings. Their Hydrogen Deficiency Index in one. The General formula for cycloalkanes is,
CnH2n
When number of Carbons = 8
Then
C₈H₂₍₈₎
C₈H₁₆
Result:
Cycloalkane containing 8 carbon atoms has
16 hydrogen atoms.
<u>Answer:</u> For the given equation, only iron has the value of
equal to 0 kJ.
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28reactant%29%5D)
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(Fe(s))})+(3\times \Delta H^o_f_{(CO_2(g))})]-[(3\times \Delta H^o_f_{(CO(g))})+(2\times \Delta H^o_f_{(Fe_2O_3(s))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28Fe%28s%29%29%7D%29%2B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%5D-%5B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO%28g%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28Fe_2O_3%28s%29%29%7D%29%5D)
The enthalpy of formation for the substances present in their elemental state is taken as 0.
Here, iron is present in its elemental state which is solid.
Hence, for the given equation, only iron has the value of
equal to 0 kJ.